16. We follow Sample Problem 40-3 in the presentation of this solution. The integration result quoted below
is discussed in a little more detail in that Sample Problem. We note that the arguments of the sine
functions used below are in radians.
L
(a) The probability of detecting the particle in the region 0 d" x d" is
4
Ą/4
Ą/4
2 L 2 y sin 2y
sin2 ydy = - =0.091 .
L Ą Ą 2 4
0
0
(b) As expected from symmetry,
Ą
Ą
2 L 2 y sin 2y
sin2 ydy = - =0.091 .
L Ą Ą 2 4
Ą/4
Ą/4
L 3L
(c) For the region d" x d" , we obtain
4 4
3Ą/4
3Ą/4
2 L 2 y sin 2y
sin2 ydy = - =0.82
L Ą Ą 2 4
Ą/4
Ą/4
which we could also have gotten by subtracting the results of part (a) and (b) from 1; that is,
1 - 2(0.091) = 0.82.